Year 12 Physics Module 5 ⏱ ~30 min 5 MC · 3 Short Answer Lesson 6 of 18

Torque and Rotational Equilibrium

A tower crane can hold a 4-tonne load hanging 30 m out along its jib using a 20-tonne concrete counterweight parked only 6 m behind the mast. The two weights are nowhere near equal, yet the crane does not topple, because what balances is not force but the turning effect of force. That turning effect, torque, and the mathematics of τ = rF sin θ are what this lesson unpacks.

Today's hook: A tower crane lifts a 4-tonne load hanging 30 m out in front of the mast, held in check by a 20-tonne counterweight only 6 m behind it. The counterweight is five times heavier but sits five times closer. Why does that exact trade-off keep the crane from rotating and collapsing, and what quantity, measured in newton metres, is actually being balanced?
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Connect and orient

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice
Before you read, predict

Why is it easier to open a door by pushing near the handle than near the hinges? What quantity is different in each case?

Warm-up: what quantity describes the turning effect of a force about a pivot?

Learning Intentions
goals

Know, Define and Calculate Torque

  • Define torque as the moment, or turning effect, of a force
  • Apply $\tau = rF_\perp = rF\sin\theta$ correctly

Understand, Analyse Lever Arms and Angles

  • Identify the perpendicular distance from pivot to line of action
  • Determine when torque is maximised or zero

Can Do, Solve Rotational Equilibrium Problems

  • Apply the principle that net torque equals zero for balance
  • Solve seesaw, door and wrench problems
Scan these before reading
vocab
Torque ($\tau$)The moment, or turning effect, of a force about a pivot. Measured in newton-metres (N·m).
Lever armThe perpendicular distance from the pivot to the line of action of the force.
Pivot (axis of rotation)The fixed point about which an object rotates.
Rotational equilibriumThe condition in which the net torque acting on an object is zero ($\sum \tau = 0$), so it does not undergo angular acceleration.
Perpendicular component ($F_\perp$)The component of force acting at right angles to the lever arm; the only component that produces torque.
Cross-lesson links: L01–L05 covered linear projectile motion, where forces produce straight-line acceleration. L06 introduces the turning effect of a force, torque (τ = rF sin θ), the quantity that decides whether an extended object rotates. It is the bridge into the circular and rotational motion block that begins with uniform circular motion in L07.
Misconceptions to fix
✗ Wrong: Torque is the same as force.
✓ Right: Torque is not a force. It is the turning effect of a force. Two equal and opposite forces can produce zero net force but a large net torque (a couple).
✗ Wrong: The distance $r$ is measured along the force direction.
✓ Right: $r$ is the distance from the pivot to the point where the force is applied, measured along the lever arm, not along the force vector.
✗ Wrong: Torque is measured in joules.
✓ Right: Torque has units of N·m; it is not energy. The unit is newton-metre, not joule.

Torque is measured in joules.

A force directed through the pivot produces a large torque.

Core Content

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Define the moment and its sign

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What is Torque?
+5 XP

The moment, or turning effect, of a force

Torque Diagram

Torque Diagram

Torque Detailed

Torque Detailed

Push a door near its hinge and it barely turns; apply the same force at the handle and it turns readily. The force is unchanged, but its moment about the hinge is larger. Torque depends on three things:

  1. The magnitude of the applied force $F$
  2. The distance $r$ from the pivot (axis of rotation) to where the force is applied
  3. The angle $\theta$ between the force vector and the lever arm
Torque Formula

$\tau = rF_\perp = rF\sin\theta$

$\tau$ = torque (N·m) · $r$ = pivot-to-application-point distance (m) · $F$ = applied force (N) · $\theta$ = angle between $\vec r$ and $\vec F$

Key insight: Only the perpendicular component of force ($F_\perp = F\sin\theta$) contributes. Equivalently, $\tau=F r_\perp$, where $r_\perp$ is the perpendicular distance from the pivot to the force’s line of action. Do not confuse $r_\perp$ with the full pivot-to-application distance $r$.

Signed torque

Declare a sign convention before adding torques. This lesson uses anticlockwise positive and clockwise negative, so $\sum\tau=\tau_1+\tau_2+\cdots$ is an algebraic sum. The opposite convention also works if used consistently.

Syllabus Requirement

The HSC syllabus uses $\tau=rF_\perp=rF\sin\theta$. You must identify $r$, $F$, $\theta$ and the turning direction in a diagram. The unit is the newton-metre (N·m); do not relabel it as joules even though the base dimensions are the same.

Torque is the moment of a force: $\tau=rF\sin\theta=Fr_\perp$ in N·m. Here $\theta$ is the angle between $\vec r$ and $\vec F$. Use anticlockwise $+$ and clockwise $-$ in signed sums. N·m has the same base dimensions as J, but torque is not energy and is never labelled J.

Pause, copy the highlighted torque formula and the unit rule into your book before moving on.

A force of 40 N is applied perpendicular to a lever 0.5 m from the pivot. The torque is:

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Represent the line of action

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Maximising and Minimising Torque
+5 XP

How angle and distance affect rotation

We just saw how torque is defined as $\tau = rF\sin\theta$. That raises a question: which angle gives the most (or least) rotation effect? This card answers it → 90° gives maximum torque; 0° or 180° gives zero.

From $\tau = rF\sin\theta$, we can identify three special cases:

Angle $\theta$ $\sin\theta$ Torque Physical situation
$90°$ (force perpendicular to lever arm) $1$ $\tau_{\max} = rF$ Pushing a door at right angles, maximum effectiveness
$0°$ or $180°$ (force parallel to lever arm) $0$ $\tau = 0$ Pushing directly toward or away from the hinge, door does not rotate
$\theta$ between $0°$ and $90°$ $0 < \sin\theta < 1$ $0 < \tau < rF$ Pushing a door at an angle, only the perpendicular component works

Two ways to increase torque:

  1. Increase the force $F$ push harder
  2. Increase the moment arm $r_\perp$, often by pushing farther from the pivot
Real World

Door handles are placed far from the hinges because this maximises $r$ and therefore maximises torque for the same pushing force. If you tried to open a door by pushing near the hinges, even a very large force would produce little rotation because $r$ is small.

Torque is maximised when $\theta = 90°$ ($\tau_{\max} = rF$) and is zero when the force is parallel to the lever arm ($\theta = 0°$ or $180°$). To increase torque: increase $F$ or increase $r$ (push farther from the pivot).

Add the highlighted angle rules to your notes before the check below.

Pushing a door perpendicular to its surface (at $90°$) produces the maximum torque.

A force directed directly toward the pivot produces a large torque.

Torque can be increased by pushing further from the pivot.

Explore the turning effect
interactive

Change the force, distance and angle. Predict the torque first, then use the calculator to test whether your sign and magnitude agree.

Open fullscreen ↗
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Worked wrench example

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Worked Example 1, Wrench on a Nut
+5 XP

Calculating torque at different angles

We just saw that torque depends on angle, with maximum at 90°. That raises a question: how do we apply this to a real calculation with different angles? This card answers it → always identify $r$, $F$ and $\theta$, then substitute into $\tau = rF\sin\theta$.

Worked example 1+5 XP on full reveal

Wrench on a nut. A mechanic applies 80 N to a 0.25 m wrench. Calculate the torque when the force is (a) perpendicular, (b) at 30° to the wrench and (c) directed along the wrench toward the nut.

1
Identify the quantities
$F=80\text{ N}$ and $r=0.25\text{ m}$.
The stated angles are measured between the wrench’s lever arm and the force.
2
Perpendicular force
$\tau=0.25(80)\sin90°=\mathbf{20\text{ N·m}}$.
This is the maximum magnitude. Its sign depends on whether the force tends to turn the nut clockwise or anticlockwise.
3
Force at 30°
$\tau=0.25(80)\sin30°=\mathbf{10\text{ N·m}}$.
Only the perpendicular force component contributes to the moment.
4
Force along the wrench
$\tau=0.25(80)\sin0°=\mathbf{0\text{ N·m}}$.
The line of action passes through the pivot, so the moment arm is zero.
5
Report magnitude and direction
The magnitudes are $20$, $10$ and $0\text{ N·m}$. Add $+$ or $-$ once the turning direction is known.
A torque answer is incomplete if a signed problem supplies a direction but the sign is omitted.
Step 1 of 5

Always identify $r$, $F$ and $\theta$ before substituting into $\tau = rF\sin\theta$. When $\theta = 0°$: $\tau = 0$ (force through the pivot); when $\theta = 90°$: $\tau = rF$ (maximum).

Pause, write the highlighted problem-solving steps into your book before the check below.

A mechanic applies 80 N to a wrench 0.25 m from the nut at 30° to the wrench handle. What is the torque?

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Worked equilibrium example

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Worked Example 2, Seesaw in Equilibrium
+5 XP

Balancing torques for zero rotation

We just saw how to calculate torque from $\tau = rF\sin\theta$. That raises a question: what happens when multiple torques act together, can they cancel? This card answers it → in rotational equilibrium $\sum\tau = 0$; clockwise torques exactly balance anticlockwise ones.

When an object is in rotational equilibrium, the sum of all torques acting on it is zero:

Rotational Equilibrium

$\sum \tau = 0$

With anticlockwise positive, clockwise torques are negative and the signed sum is zero.

Worked example 2+5 XP on full reveal

Seesaw in equilibrium. A uniform 3.0 m seesaw is pivoted at its centre. A 30 kg child sits 1.2 m left of the pivot. Where must a 25 kg child sit on the right to balance it?

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Choose signs and a pivot
Take anticlockwise as $+$ and moments about the central pivot.
The pivot reaction then has zero moment because its line of action passes through the pivot.
2
Write the signed sum
$\sum\tau=(30g)(1.2)-(25g)r_2=0$.
The children’s weights act on opposite sides and produce opposite turning tendencies.
3
Solve
$r_2=\dfrac{30g(1.2)}{25g}=\mathbf{1.44\text{ m}}$.
$g$ cancels because both forces are weights in the same gravitational field.
4
Check feasibility
The right half is only $1.50\text{ m}$ long, and $1.44\text{ m}<1.50\text{ m}$.
A numerical answer outside the physical seesaw would reveal an impossible arrangement.
5
State the result
The 25 kg child sits $\mathbf{1.44\text{ m}}$ to the right of the pivot.
For complete static equilibrium, the support force must also make $\sum F_y=0$; the moment calculation alone establishes rotational equilibrium.
Step 1 of 5

Rotational equilibrium: $\sum\tau = 0$ (clockwise torques = anticlockwise torques). For two weights on a balanced beam: $r_1 m_1 g = r_2 m_2 g$, so $r_1 m_1 = r_2 m_2$ ($g$ cancels). Heavier objects sit closer to the pivot.

Add the highlighted equilibrium condition to your notes before the check below.

A 30 kg child sits 1.2 m from the pivot on a seesaw. A 25 kg child sits on the other side. What distance from the pivot balances the seesaw?

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Apply, diagnose and consolidate

Beyond the syllabus. Applied torque and rotational equilibrium are core. Moment of inertia, angular momentum and rigid-body dynamics beyond torque balance are extension — exam questions stay with torque and balancing torques.
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Scope Boundary, What Comes Next
extension

Keep assessed statics separate from later rotational dynamics

Assessed in this lesson

Define and calculate $\tau=rF\sin\theta=Fr_\perp$; identify force angle and line of action; use a declared clockwise/anticlockwise sign convention; solve $\sum\tau=0$; and recognise that complete static equilibrium requires both $\sum F=0$ and $\sum\tau=0$.

Extension, not assessed here

The relationship $\tau_{\text{net}}=I\alpha$ belongs to rotational dynamics. It explains how a net torque changes angular velocity, but you are not required to calculate with moment of inertia $I$ or angular acceleration $\alpha$ in this lesson.

Extension only

$\tau_{\text{net}} = I\alpha$

net torque = moment of inertia × angular acceleration

Assessment boundary: learn $\tau=rF\sin\theta$, signed torque sums, $\sum\tau=0$, and the two conditions for complete static equilibrium. Treat $\tau_{\text{net}}=I\alpha$ as extension context only.

Copy the assessed list; the extension equation is optional.

Three of these statements about torque are correct. Pick the odd one out.

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Common Misconceptions
+5 XP

Errors to avoid in torque problems

We just set the assessment boundary. Now diagnose the common errors: torque is not a force, $r$ and $r_\perp$ are different distances, and the unit is N·m rather than J.

✗ "Torque is a force"
✓ Torque is not a force. It is the turning effect of a force. Two equal and opposite forces can produce zero net force but a large net torque (a couple).
✗ "The distance $r$ is along the force direction"
✓ In $\tau=rF\sin\theta$, $r$ runs from the pivot to the application point. In $\tau=Fr_\perp$, $r_\perp$ is the perpendicular distance to the force’s line of action.
✗ "Torque is measured in joules"
✓ Torque has units of N·m; it is not energy. The unit is called a newton-metre, not a joule.

Three key torque rules: (1) torque ≠ force, it is the turning effect; (2) distinguish pivot-to-point distance $r$ from moment arm $r_\perp$; (3) units are N·m, never J, even though both have dimensions kg·m²·s⁻².

Pause, write the highlighted unit rule and the three distinctions into your book before moving on.

Complete the sentence. For rotational equilibrium, the sum of all torques must equal _____.

Activity 1, Torque Calculations
ApplyBand 3

Practise calculating torque and rotational equilibrium.

  1. A force of 40 N is applied perpendicular to a spanner, 0.25 m from the bolt. Calculate the torque about the bolt.
  2. A 60 N force is applied to a lever at 30° to the lever arm, 0.50 m from the pivot. Find the torque ($\tau = rF\sin\theta$).
  3. On a see-saw, a 30 kg child sits 1.5 m to the left of the pivot. How far to the right must a 45 kg child sit to balance it (rotational equilibrium)?

A 50 N force is applied perpendicular to a spanner, 0.20 m from the nut. The torque about the nut is:

Activity 2, Why Levers Work
UnderstandBand 4

Explain torque in terms of force and lever arm.

Explain why using a longer spanner makes it easier to undo a tight bolt, even when you push with the same force. Use the relationship $\tau = rF\sin\theta$ in your answer, and explain why pushing at 90° to the spanner is most effective.

Wrap-up, Misconceptions & Summary

Misconceptions, final check

✗ "A force always produces rotation if it is applied away from the pivot."
✓ From $\tau = rF\sin\theta$, torque depends on the angle $\theta$. When $\theta = 0°$ (force directed along the lever arm toward the pivot), $\sin 0° = 0$ and $\tau = 0$, even for a large force applied far from the pivot.
✗ "Torque has units of joules since J = N·m."
✓ Although N·m and joules have the same dimensions, they are conceptually distinct. Torque is not energy and must always be expressed in N·m, never in J.

Copy into your books

Key Definitions

  • Torque: moment, or turning effect, of a force
  • Lever arm: perpendicular distance from pivot to line of force
  • Rotational equilibrium: $\sum\tau = 0$
  • Static equilibrium: $\sum F=0$ and $\sum\tau=0$

Torque Formula

  • $\tau = rF_\perp = rF\sin\theta$
  • Units: N·m (not joules)
  • Max at $\theta = 90°$; zero at $\theta = 0°$

Equilibrium Rule

  • Choose anticlockwise $+$, clockwise $-$
  • $\sum\tau = 0$ for rotational balance
  • $r_1 m_1 = r_2 m_2$ for seesaws
  • $g$ cancels in balance problems

Key Principles

  • Only $F\sin\theta$ causes rotation
  • More distance from pivot → more torque
  • Always distinguish $r$ from $r_\perp$

Which set of statements correctly describes torque?

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Apply torque independently

Quick recall, torque and rotational motion
+5 XP

A fresh five-question set drawn from this lesson's bank, feedback shown immediately. +5 XP per correct · +25 XP all correct

Pick your answer, then rate your confidence, that tells the system what to drill next.

Short Answer, 12 marks
+5 XP

ApplyBand 4(3 marks) 1. A force of 50 N is applied to a door handle 0.80 m from the hinges. Calculate the torque magnitude when the force is (a) perpendicular to the lever arm, (b) at 60° to the lever arm. Explain which pushing direction is most effective for opening the door.

1 mark: correct torque for (a) · 1 mark: correct torque for (b) · 1 mark: explanation using $\sin\theta$

AnalyseBand 5(4 marks) 2. A metre ruler is pivoted at the 50 cm mark. A 200 g mass is hung at the 20 cm mark. Where must a 150 g mass be placed to balance the ruler horizontally? Show all working and state the principle used.

1 mark: principle stated · 1 mark: correct $r_1$ from pivot · 1 mark: correct method · 1 mark: position on ruler

EvaluateBand 6(5 marks) 3. Evaluate the statement: "A force always produces rotation if it is applied away from the pivot." Use the torque formula to identify a situation where a non-zero force applied away from the pivot produces zero torque. Explain the physical meaning of this result.

1 mark: identifies statement as incorrect · 1 mark: references $\tau = rF\sin\theta$ · 1 mark: identifies $\theta = 0°$ as the zero-torque case · 1 mark: physical explanation · 1 mark: example

Show all answers

Multiple choice

MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.

Short Answer, Model Answers

Q1 (3 marks):

(a) $\tau = rF\sin\theta = 0.80 \times 50 \times \sin(90°) = 0.80 \times 50 \times 1 = \mathbf{40\ \text{N·m}}$ (1 mark)

(b) $\tau = 0.80 \times 50 \times \sin(60°) = 0.80 \times 50 \times 0.866 = \mathbf{34.6\ \text{N·m}}$ (1 mark)

The perpendicular push is most effective because $\sin(90°) = 1$ is the maximum value of $\sin\theta$. Any angle less than $90°$ reduces the perpendicular component and therefore reduces torque (1 mark).

Q2 (4 marks):

Principle: For rotational equilibrium, the sum of clockwise torques equals the sum of anticlockwise torques ($\sum \tau = 0$) (1 mark).

Distance of 200 g mass from pivot: $r_1 = 50 - 20 = 30\ \text{cm} = 0.30\ \text{m}$

Let $r_2$ be the distance of the 150 g mass on the opposite side.

$r_1 m_1 g = r_2 m_2 g \Rightarrow r_2 = \dfrac{r_1 m_1}{m_2} = \dfrac{0.30 \times 0.200}{0.150} = \dfrac{0.060}{0.150} = \mathbf{0.40\ \text{m}}$ (2 marks for method + answer)

Position: $50 + 40 = \mathbf{90\ \text{cm}}$ mark (or 40 cm to the right of the pivot) (1 mark).

Q3 (5 marks):

The statement is incorrect (1 mark). From $\tau = rF\sin\theta$, torque depends on the angle $\theta$ between the force and the lever arm, not just on the magnitude of the force or its distance from the pivot (1 mark).

When $\theta = 0°$ or $180°$ (force directed along the lever arm, either toward or away from the pivot), $\sin\theta = 0$ and therefore $\tau = 0$ (1 mark).

Physical meaning: A force directed through or away from the pivot pulls along the lever arm but does not tend to rotate the object (1 mark). For example, pushing a door directly toward its hinges produces no rotation regardless of how hard you push (1 mark).

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Retrieve and reflect

Check what actually stuck
Take the full module quiz
quiz

A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.

Start the module quiz →
How did your thinking change?

At the start you were asked why a 20-tonne counterweight parked 6 m behind a crane's mast can hold up a 4-tonne load hanging 30 m out in front.

You can now answer it with torque. The load's torque about the mast is $\tau = rF\sin\theta = 30 \times (4000 \times 9.8) \times \sin 90° \approx 1.2 \times 10^{6}$ N·m clockwise. The counterweight's torque is $6 \times (20000 \times 9.8) \approx 1.2 \times 10^{6}$ N·m anticlockwise. The forces differ by a factor of five, but the lever arms differ by the same factor in the opposite sense, so the net torque is zero. The mast and ground must also supply support forces so that $\sum F=0$; only when both force and torque conditions hold is the stationary crane in complete static equilibrium.

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